On Wikipedia:
"Atomic physics
Mills says that the electron is an extended particle which in free space
is
a flat disk of spinning charge[citation needed]. His new model treats the
electron, not as a point nor as a probability wave, but as a dynamic
two-dimensional spherical shell surrounding the nucleus. The resulting
model, called the "orbitsphere", provides a fully classical physical
explanation for phenomena such as quantization, angular momentum, Bohr
magneton. Essentially, the electron orbitsphere is a "dynamic spherical
resonator cavity" that traps photons of discrete frequencies."
I remember reading in Feynman vol. 2 that when this kind of model was
tried
it lead to inconsistencies such as different parts of the electron having
relative motions greater than "c".
Has he done any direct measurement of this? What is the cross-section of
free electrons (not my area is PP)? What can he do with the simplest set
up
to prove this conjecture?
I mean that's how research gets done in main universities. It's slow and
frustrating but the good people pass through the system eventually. One
dots
the "Is" crosses the "Ts".
What if he has something producing excess heat and then clouds it all with
stuff people can't begin to digest?
-----Original Message-----
From: Mike Carrell [mailto:[EMAIL PROTECTED]
Sent: 25 October 2008 22:05
To: vortex-l@eskimo.com
Subject: Re: [Vo]:Banking on BLP?
Ed wrote:
Subject: Re: [Vo]:Banking on BLP?
Robin, my main point is that an electron leaving an atom cannot go to
infinity under the conditions Mills has in his reactor. At most, it
will
go into some other energy level, such as the conduction band if one
exists in the material. This fact is not based on speculation,
assumptions, or theory. This is a simple fact of nature that is well
understood.
MC: Which values and which electrons, Ed? In eq. 23, two electrons a
'liberated' to facilitate catalyzing H[1/3]. The physical situation in the
cell is NaH resident within the R-Ni mesh, which has an enormous surface
area. On the scale of a molecule, why can't the electrons wander away?
There
are He atoms at 760 Torr hanging around too. The electron bound to the
catalyzed H doesn't go anywhere, it just gets closer to its proton. Now I
don't yet understand where the energy to ionize the Na comes from, but the
DSC plot shows *something* happens. *That* requires eplanation.
The values Mills uses to evaluate the process are all based on the
electron going to infinity. Therefore, these values simply cannot apply
to the real process. Instead, Mills assumes an unrealistic process to
make his numbers fit his expectation.
MC: Are you also including the ionic catalysts in the gas phase cells?
If we accept the excess power he claims, the process must be different
from the one he proposes.
MC: Why so? These solid fuel cells are a continuum with years of work in
the
electrolytic and gas phases. There are dozens of reports and papers
supporting lthe reactions. Good calorimetry has been done iwth microwace
excitation by Jonatan Phillips at the University of New Mexico. He was in
town during ICCF-14 and slipped in to put up a poster on his calorimetric
studies. In an early version of his reports there is a statement that the
heat measured implied substantial conversion to H[1/4]. Philipps is
currentlyas Distinguished Professor at the Farris center, supported by Los
Alamos. He has a long association with Mills. I very strongly suggest that
you contact him; he may be very helpful.
This is important to me, because I'm trying
to identify the Mills catalyst that is making hydrinos in the CF
process,
which has similar restrictions.
MC: H and D atoms can autocatalyze in a three-body reaction because 2H+
provide the 27.2 energy for catalysis. Because it is a three-body
reaction,
the reaction density is low but favored by H and D rich environments such
as
the LENR environments. A reactions density too low for optical observation
may yet be very intense on the particle-counting scene.
An assumption on his part
that is unrealistic and impossible does me no good in trying to use his
method in this search. Therefore, I'm trying to understand what is
actually happening in his cell because the hydrino process appears to be
real under these conditions. Only his explanation makes no sense.
MC: Granted, there are problems, as with LENR phenomena which don't make
sense either. Nature is trying to tell us something.
Mike Carrell
Regards,
Ed
On Oct 24, 2008, at 9:47 PM, Robin van Spaandonk wrote:
In reply to Edmund Storms's message of Fri, 24 Oct 2008
6:05:50 -0600:
Hi,
[snip]
I think you are close to describing the process, Robin. Simply
decomposing NaH cannot result in hydrinos because the expected ion is
not formed.
Absence of evidence is not evidence of absence, unless someone
explicitly looked
for it under the right conditions, and didn't find it.
On the other hand, as you suggest, if the decomposition
occurs on the Ni surface, the Na will have a complex ion state because
it now is an absorbed atom, not a free, isolated atom. In addition,
the electron that is promoted to a higher level has a place to go,
i.e. into the conduction band of the Ni. The only problem is
achieving a match between the energy change of the promoted electron
and the energy shrinkage of the hydrino electron.
I suspect you are needlessly multiplying entities. ;)
IOW Mills provides a catalyst that has the necessary property, and gets
the
expected result. Why is it so hard to accept that he might be right?
Granted spectroscopic results indicating presence of Na++ would go a
long way to
proving him right.
Now for a question. Why must the electron that is promoted always
come from a level that is observed to form an ion during normal
ionization?
Personally, I don't think it does, and have previously suggested that
Li, which
has an x-ray absorption energy of 54.75 eV, may be an example of this.
However
Na doesn't appear to fit the bill.
For example, removal of a 2p electron from Na++ would
occur during "normal" ionization, but is this happening here?
No, but then Na++ is not the catalyst either. The whole molecule is the
catalyst. BTW the third ionization energy of Na is 71.641 eV, and none
of the
immediate reactions have enough energy to do this. Only a further
reaction of
H[1/3] to a lower level would provide such energy. (3->4 yields 95 eV).
In
other words, why can't a 1s electron be removed from a neutral Na
without the 2p electron being affected. After the 1s electron is
removed, a 2p electron would take its place and release a small
amount of energy as X-rays. This energy would be a byproduct of the
process just like the hydrino energy.
Do you know how much energy is required to remove a 1s electron from
nearly neutral Na?
1073 eV. (K shell x-ray absorption energy).
The process gets more unknown because the electron
would be promoted into the conduction band, which has a lower energy
than vacuum. In other words, perhaps Mills has the right process but
is using the wrong electron promotion process to describe it simply
because the wrong promotion gives the expected energy.
If so, then I think you need to come up with an alternative (and the
numbers to
back it up). The work function of the metal might be a good place to
start,
however in this case we're looking at an alloy/compound, which
complicates
matters.
[snip]
Regards,
Robin van Spaandonk <[EMAIL PROTECTED]>
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